đây nè bạn \(A=\left(\sqrt{a}+\dfrac{1}{\sqrt{a}}\right)\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{2\sqrt{a}}{a\sqrt{a}+\sqrt{a}-a-1}\right)\)
\(a.A=\left(\sqrt{a}+\dfrac{1}{\sqrt{a}}\right)\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{2\sqrt{a}}{a\sqrt{a}+\sqrt{a}-a-1}\right)=\dfrac{a+1}{\sqrt{a}}.\dfrac{a+1-2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}-1\right)}=\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}\left(\sqrt{a}-1\right)}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\) ( a > 0 ; a # 1 )
\(b.A^2=\dfrac{1}{4}\text{ ⇔}\dfrac{a-2\sqrt{a}+1}{a}-\dfrac{1}{4}=0\)
⇔ \(\dfrac{4a-8\sqrt{a}+4-a}{4\sqrt{a}}=0\)
⇔ \(3a-2\sqrt{a}-6\sqrt{a}+4=0\)
⇔ \(\left(\sqrt{a}-2\right)\left(3\sqrt{a}-2\right)=0\)
⇔ \(a=4\left(TM\right)orx=\dfrac{4}{9}\left(TM\right)\)
KL............
